YES
We show the termination of the TRS R:

  app(app(map(),f),nil()) -> nil()
  app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs))
  app(app(minus(),x),|0|()) -> x
  app(app(minus(),app(s(),x)),app(s(),y)) -> app(app(minus(),x),y)
  app(app(div(),|0|()),app(s(),y)) -> |0|()
  app(app(div(),app(s(),x)),app(s(),y)) -> app(s(),app(app(div(),app(app(minus(),x),y)),app(s(),y)))

Instead we prove termination of the uncurried system:

  app(cons_0(),y) -> cons_1(y)
  app(cons_1(x1),y) -> cons_2(x1,y)
  app(div_0(),y) -> div_1(y)
  app(div_1(x1),y) -> div_2(x1,y)
  app(map_0(),y) -> map_1(y)
  app(map_1(x1),y) -> map_2(x1,y)
  app(minus_0(),y) -> minus_1(y)
  app(minus_1(x1),y) -> minus_2(x1,y)
  app(s_0(),y) -> s_1(y)
  map_2(f,nil_0()) -> nil_0()
  map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
  minus_2(x,|0|_0()) -> x
  minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
  div_2(|0|_0(),s_1(y)) -> |0|_0()
  div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: app#(div_1(x1),y) -> div_2#(x1,y)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: app#(minus_1(x1),y) -> minus_2#(x1,y)
p4: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p5: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)
p6: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)
p7: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(y))
p8: div_2#(s_1(x),s_1(y)) -> minus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The estimated dependency graph contains the following SCCs:

  {p2, p4, p5}
  {p7}
  {p6}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: app#(map_1(x1),y) -> map_2#(x1,y)
p3: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r1, r2, r3, r7, r9, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = x1 + 6
        cons_2_A(x1,x2) = max{3, x1 - 22, x2 - 22}
        app#_A(x1,x2) = x1 + 5
        map_1_A(x1) = max{22, x1 + 7}
        app_A(x1,x2) = max{21, x1 - 8, x2 - 12}
        cons_0_A = 1
        cons_1_A(x1) = max{2, x1 - 13}
        div_0_A = 6
        div_1_A(x1) = max{7, x1 - 13}
        minus_0_A = 7
        minus_1_A(x1) = max{6, x1 - 26}
        s_0_A = 9
        s_1_A(x1) = max{10, x1 - 12}
        div_2_A(x1,x2) = max{11, x1 - 22, x2 - 45}
        |0|_0_A = 0
  
    base order 2:
  
      precedence:
      
        map_1 > app# = app > cons_2 = s_1 = div_2 > map_2# = cons_1 = div_1 = s_0 = |0|_0 > cons_0 = div_0 = minus_1 > minus_0
      
  
    partial status:
  
      pi(map_2#) = [1]
      pi(cons_2) = []
      pi(app#) = [1]
      pi(map_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(div_0) = []
      pi(div_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(div_2) = []
      pi(|0|_0) = []

The following rules are removed:

  p2



-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: map_2#(f,cons_2(x,xs)) -> app#(f,x)
p2: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r1, r3, r5, r9, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        map_2#_A(x1,x2) = max{x1 + 2, x2 + 3}
        cons_2_A(x1,x2) = max{x1 - 1, x2}
        app_A(x1,x2) = max{0, x1 - 4, x2 - 3}
        cons_0_A = 1
        cons_1_A(x1) = max{0, x1 - 4}
        div_0_A = 3
        div_1_A(x1) = max{0, x1 - 3}
        map_0_A = 3
        map_1_A(x1) = max{0, x1 - 3}
        s_0_A = 1
        s_1_A(x1) = max{0, x1 - 3}
        div_2_A(x1,x2) = max{1, x1 - 5, x2 - 9}
        |0|_0_A = 0
  
    base order 2:
  
      precedence:
      
        app > map_2# = cons_2 = div_0 = map_1 = s_1 = div_2 > cons_1 = div_1 = map_0 = s_0 = |0|_0 > cons_0
      
  
    partial status:
  
      pi(map_2#) = [2]
      pi(cons_2) = [2]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(div_0) = []
      pi(div_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(s_0) = []
      pi(s_1) = []
      pi(div_2) = []
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: div_2#(s_1(x),s_1(y)) -> div_2#(minus_2(x,y),s_1(y))

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r1, r2, r3, r5, r7, r12, r13

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        div_2#_A(x1,x2) = max{1, x1 - 12}
        s_1_A(x1) = max{19, x1 + 14}
        minus_2_A(x1,x2) = max{11, x1}
        app_A(x1,x2) = max{4, x1 - 5, x2 - 3}
        cons_0_A = 3
        cons_1_A(x1) = max{4, x1 - 4}
        cons_2_A(x1,x2) = max{1, x1 - 10, x2 - 17}
        div_0_A = 1
        div_1_A(x1) = max{2, x1 - 14}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 10}
        minus_0_A = 4
        minus_1_A(x1) = max{0, x1 - 4}
        |0|_0_A = 18
  
    base order 2:
  
      precedence:
      
        minus_2 > div_2# = s_1 > app = cons_0 = |0|_0 > div_1 = minus_1 > cons_2 = div_0 = map_1 = minus_0 > cons_1 = map_0
      
  
    partial status:
  
      pi(div_2#) = []
      pi(s_1) = []
      pi(minus_2) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(div_0) = []
      pi(div_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus_2#(s_1(x),s_1(y)) -> minus_2#(x,y)

and R consists of:

r1: app(cons_0(),y) -> cons_1(y)
r2: app(cons_1(x1),y) -> cons_2(x1,y)
r3: app(div_0(),y) -> div_1(y)
r4: app(div_1(x1),y) -> div_2(x1,y)
r5: app(map_0(),y) -> map_1(y)
r6: app(map_1(x1),y) -> map_2(x1,y)
r7: app(minus_0(),y) -> minus_1(y)
r8: app(minus_1(x1),y) -> minus_2(x1,y)
r9: app(s_0(),y) -> s_1(y)
r10: map_2(f,nil_0()) -> nil_0()
r11: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs))
r12: minus_2(x,|0|_0()) -> x
r13: minus_2(s_1(x),s_1(y)) -> minus_2(x,y)
r14: div_2(|0|_0(),s_1(y)) -> |0|_0()
r15: div_2(s_1(x),s_1(y)) -> s_1(div_2(minus_2(x,y),s_1(y)))

The set of usable rules consists of

  r1, r2, r3, r5, r7, r14

Take the reduction pair:

  generalized weighted path order (>=MG, >MG)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus_2#_A(x1,x2) = max{x1 + 2, x2 + 9}
        s_1_A(x1) = max{12, x1 + 8}
        app_A(x1,x2) = max{11, x1 - 3, x2 - 5}
        cons_0_A = 2
        cons_1_A(x1) = max{1, x1 - 5}
        cons_2_A(x1,x2) = max{2, x1 - 12, x2 - 19}
        div_0_A = 1
        div_1_A(x1) = max{2, x1 - 10}
        map_0_A = 1
        map_1_A(x1) = max{2, x1 - 16}
        minus_0_A = 2
        minus_1_A(x1) = max{1, x1 - 6}
        div_2_A(x1,x2) = max{10, x1 - 14, x2 - 19}
        |0|_0_A = 9
  
    base order 2:
  
      precedence:
      
        div_2 > minus_2# = s_1 > app = cons_0 = minus_0 = |0|_0 > cons_2 = div_1 = map_1 = minus_1 > cons_1 = div_0 = map_0
      
  
    partial status:
  
      pi(minus_2#) = [2]
      pi(s_1) = [1]
      pi(app) = []
      pi(cons_0) = []
      pi(cons_1) = []
      pi(cons_2) = []
      pi(div_0) = []
      pi(div_1) = []
      pi(map_0) = []
      pi(map_1) = []
      pi(minus_0) = []
      pi(minus_1) = []
      pi(div_2) = []
      pi(|0|_0) = []

The following rules are removed:

  p1

Then no dependency pair remains.